I'm stuck on this problem:
Given $n$ complex numbers $\alpha_\nu$, show that the determinant of
$$ \begin{vmatrix} \alpha_1 & \alpha_2 & \cdots & \alpha_{n-1} & \alpha_n \\ \alpha_2 & \alpha_3 & \cdots & \alpha_n & \alpha_1 \\ \vdots \\ \alpha_n & \alpha_1 & \cdots & \alpha_{n-2} & \alpha_{n-1} \end{vmatrix} $$
is equivalent to $(-1)^{\frac{n(n-1)}{2}}\beta\cdots\beta_n$ where $$\beta_k=\sum_\nu \epsilon^\nu_k \alpha_\nu$$ and $$\epsilon_k=cos(\frac{2\pi k}{n})+isin(\frac{2\pi k}{n})$$.
Hint: Multiply the above matrix by $$ \begin{pmatrix} \epsilon_1 & \cdots & \epsilon_n\\ \vdots\\ \epsilon_1^n & \cdots & \epsilon_n^n \end{pmatrix} $$
I didn't know exactly how to multiply out that matrix, but I think I may have found another way to solve this. So, what I've done is first notice that this matrix permutes backwards every row down. I wanted to proceed by induction. I let my base case be $n=2$ and I came up with
$$ \begin{vmatrix} \alpha_1 & \alpha_2\\ \alpha_2 & \alpha_1 \end{vmatrix} =\alpha_1^2-\alpha_2^2 $$
Then, for $n=2$, I computed $\beta_1$ and $\beta_2$ like so: $$\beta_1=\epsilon_1^1 \alpha_1+\epsilon_1^2 \alpha_2=-\alpha_1+\alpha_2$$ and $$\beta_2=\epsilon_2^1 \alpha_1+\epsilon_2^2 \alpha_2=\alpha_1+\alpha_2$$.
When I plug in $n=2$ into our equivalence term, I see that this is just $-1 \beta_1 \beta_2$. So this works.
I have no clue how to even start the induction step. So here are my questions: 1. Could someone possibly give me a hint on how to start the induction step on this problem? We are using leibniz notation for determinants. 2. How do I multiply this matrix out like the hint says? 3. Would multiplying this matrix out make a big difference? Any help is greatly appreciated! Thanks!